Arithmetic Operations Questions and Answers

Answer: C

Correct answer: \( \frac{1}{5} \)

Write each factor as a single fraction: \[ 1 - \frac{1}{n} = \frac{n-1}{n}. \] The product becomes \[ \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \cdots \times \frac{9}{10}. \] Every numerator cancels against the denominator of the previous factor - a telescoping product - leaving only the first numerator and the last denominator: \[ \frac{2}{10} = \frac{1}{5}. \] The general result is that \[ \prod_{n=k}^{m}\left(1 - \frac{1}{n}\right) = \frac{k-1}{m}, \] which here is \( 2/10 \). Recognising the cancellation is the whole of the work; multiplying the eight fractions out is slow and prone to error. The same structure appears with \( \left(1 + \frac{1}{n}\right) \) factors, which give \( \frac{m+1}{k} \) instead - the fraction inverts, so it is worth checking which form is presented.

Answer: C

Correct answer: 64

Find the largest perfect square not exceeding 2,000. Since \( 44^{2} = 1936 \) and \( 45^{2} = 2025 \), the largest is 1,936. The number to subtract is therefore \[ 2000 - 1936 = 64. \] The companion question - the least number to be added to make a perfect square - uses the next square up: \( 2025 - 2000 = 25 \). Reading which is asked decides the answer, and the two are always different. The method is to take the square root, discard the fractional part for subtraction or round up for addition, and square the result. Here \( \sqrt{2000} \approx 44.7 \), so 44 and 45 are the two squares to consider. The distractor 44 is the square root itself rather than the difference, which is the usual slip. Knowing the squares up to 50 by heart makes questions of this kind almost immediate.

Answer: C

Correct answer: 5

Express 32 as a power of 2: \[ 32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^{5}. \] Since the bases are equal, the exponents must be equal, so \( x = 5 \). The rule being used is that \( a^{m} = a^{n} \) implies \( m = n \), provided \( a \) is positive and not 1. The powers of 2 are worth knowing by heart, since they appear constantly in both aptitude and computing questions: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024. The laws of indices that go with this are the other things to have ready: \[ a^{m} \times a^{n} = a^{m+n}, \quad \frac{a^{m}}{a^{n}} = a^{m-n}, \quad (a^{m})^{n} = a^{mn}, \quad a^{0} = 1, \quad a^{-n} = \frac{1}{a^{n}}. \] The distractor 16 is 32 halved, offered for a reader who confuses the exponent with the value.

Answer: A

Correct answer: 65

A number leaving the same remainder 5 on division by each divisor is 5 more than a common multiple of them all. The smallest such multiple is the LCM: \( 12 = 2^{2} \times 3 \), \( 15 = 3 \times 5 \), \( 20 = 2^{2} \times 5 \), so \[ \text{LCM} = 2^{2} \times 3 \times 5 = 60. \] The required number is therefore \[ 60 + 5 = 65. \] Checking: 65 divided by 12 gives remainder 5, by 15 gives 5, and by 20 gives 5. The general form is \( \text{LCM} \times n + r \), so the next such numbers are 125, 185 and so on. A variation asks for a number leaving different remainders whose difference from the divisor is constant - say remainders 7, 10 and 15 for divisors 12, 15 and 20, each 5 short. The answer is then \( \text{LCM} - 5 = 55 \), and recognising which form is being asked is the whole difficulty.

Answer: B

Correct answer: 0.13

A square root halves the number of decimal places, and \( \sqrt{169} = 13 \). Since 0.0169 has four decimal places, the root has two: \[ \sqrt{0.0169} = 0.13. \] Written as a fraction the reasoning is plain: \[ \sqrt{0.0169} = \sqrt{\frac{169}{10000}} = \frac{13}{100} = 0.13. \] Checking is immediate: \( 0.13 \times 0.13 = 0.0169 \). The rule requires an even number of decimal places for the root to terminate, which is why such questions always supply two, four or six. A cube root divides the places by three instead, so \( \sqrt[3]{0.000027} = 0.03 \). The perfect squares up to 30 are worth knowing by heart, since almost every question of this kind is built on one: 121, 144, 169, 196, 225, 256, 289, 324, 361, 400.

Answer: B

We can multiply sequentially.

\(0.5 \times 0.05 = 0.025\).

Now, \(0.025 \times 500 = 25 \times 0.5 = 12.5\).

Alternatively, write as fractions: \((\frac{1}{2}) \times (\frac{5}{100}) \times 500 = (\frac{1}{2}) \times (\frac{1}{20}) \times 500 = \frac{500}{40} = \frac{50}{4} = 12.5\).

Answer: C

Correct answer: 1

Look at the remainders of successive powers of 3 on division by 5: \[ 3^{1} = 3,\quad 3^{2} = 9 \equiv 4,\quad 3^{3} = 27 \equiv 2,\quad 3^{4} = 81 \equiv 1. \] The pattern 3, 4, 2, 1 then repeats, so the cycle length is 4. Since \( 40 \) is exactly divisible by 4, \( 3^{40} \) corresponds to the last term of the cycle, and the remainder is 1. A remainder of 0 in the exponent division always means the last term, not the first - which is the commonest slip in these questions. The result also follows from the unit digit: the unit digits of powers of 3 cycle 3, 9, 7, 1, so \( 3^{40} \) ends in 1, and a number ending in 1 leaves remainder 1 on division by 5. For a prime divisor Fermat’s little theorem gives the same conclusion directly: \( 3^{4} \equiv 1 \pmod 5 \).

Answer: B

Correct answer: 3

The unit digits of the powers of 7 repeat in a cycle of four: \[ 7^{1} = 7,\quad 7^{2} = 49,\quad 7^{3} = 343,\quad 7^{4} = 2401, \] giving the sequence 7, 9, 3, 1 and then repeating. To find where the 95th power falls, divide the exponent by the cycle length: \( 95 \div 4 \) leaves a remainder of 3. The third term of the cycle is 3. A remainder of 0 would mean the last term of the cycle, not the first - that is the usual trap. The cycle lengths for the other digits are worth memorising: 2, 3, 7 and 8 all cycle in four; 4 and 9 cycle in two; and 0, 1, 5 and 6 always end in themselves. The same technique answers questions about the unit digit of a product or sum of large powers: find each unit digit separately, then combine them.

Answer: B

Work outward from the inner percentage.

\[ 50\% \text{ of } 200 = 100 \]

\[ 25\% \text{ of } 100 = 25 \]

Equivalently \( 0.25 \times 0.5 \times 200 = 25 \); the two percentages multiply rather than add.

Answer: B

Correct answer: 2

Divide 1,056 by 23: \( 23 \times 45 = 1035 \), leaving a remainder of \( 1056 - 1035 = 21 \). To reach the next multiple of 23, add the difference between the divisor and the remainder: \[ 23 - 21 = 2. \] Checking, \( 1056 + 2 = 1058 = 23 \times 46 \). The distractor 21 is the remainder itself, and it is the answer to the opposite question - the least number to be subtracted to make the figure divisible. Reading which is asked is where marks are usually lost. The two rules are worth stating together. To make a number divisible by \( d \), subtract the remainder \( r \), or add \( d - r \). If the remainder is zero the number is already divisible and nothing need be added. The same reasoning extends to finding the nearest multiple in either direction.